.TH  DPFTRS 1 "November 2008" " LAPACK routine (version 3.2)                                    " " LAPACK routine (version 3.2)                                    " 
.SH NAME
DPFTRS - solves a system of linear equations A*X = B with a symmetric positive definite matrix A using the Cholesky factorization A = U**T*U or A = L*L**T computed by DPFTRF
.SH SYNOPSIS
.TP 19
SUBROUTINE DPFTRS(
TRANSR, UPLO, N, NRHS, A, B, LDB, INFO )
.TP 19
.ti +4
CHARACTER
TRANSR, UPLO
.TP 19
.ti +4
INTEGER
INFO, LDB, N, NRHS
.TP 19
.ti +4
DOUBLE
PRECISION A( 0: * ), B( LDB, * )
.SH PURPOSE
DPFTRS solves a system of linear equations A*X = B with a symmetric
positive definite matrix A using the Cholesky factorization
A = U**T*U or A = L*L**T computed by DPFTRF.
.SH ARGUMENTS
.TP 10
TRANSR    (input) CHARACTER
= \(aqN\(aq:  The Normal TRANSR of RFP A is stored;
.br
= \(aqT\(aq:  The Transpose TRANSR of RFP A is stored.
.TP 8
UPLO    (input) CHARACTER
.br
= \(aqU\(aq:  Upper triangle of RFP A is stored;
.br
= \(aqL\(aq:  Lower triangle of RFP A is stored.
.TP 8
N       (input) INTEGER
The order of the matrix A.  N >= 0.
.TP 8
NRHS    (input) INTEGER
The number of right hand sides, i.e., the number of columns
of the matrix B.  NRHS >= 0.
.TP 8
A       (input) DOUBLE PRECISION array, dimension ( N*(N+1)/2 ).
The triangular factor U or L from the Cholesky factorization
of RFP A = U**T*U or RFP A = L*L**T, as computed by DPFTRF.
See note below for more details about RFP A.
.TP 8
B       (input/output) DOUBLE PRECISION array, dimension (LDB,NRHS)
On entry, the right hand side matrix B.
On exit, the solution matrix X.
.TP 8
LDB     (input) INTEGER
The leading dimension of the array B.  LDB >= max(1,N).
.TP 8
INFO    (output) INTEGER
= 0:  successful exit
.br
< 0:  if INFO = -i, the i-th argument had an illegal value
.SH FURTHER DETAILS
We first consider Rectangular Full Packed (RFP) Format when N is
even. We give an example where N = 6.
.br
    AP is Upper             AP is Lower
.br
 00 01 02 03 04 05       00
.br
    11 12 13 14 15       10 11
.br
       22 23 24 25       20 21 22
.br
          33 34 35       30 31 32 33
.br
             44 45       40 41 42 43 44
.br
                55       50 51 52 53 54 55
.br
Let TRANSR = \(aqN\(aq. RFP holds AP as follows:
.br
For UPLO = \(aqU\(aq the upper trapezoid A(0:5,0:2) consists of the last
three columns of AP upper. The lower triangle A(4:6,0:2) consists of
the transpose of the first three columns of AP upper.
.br
For UPLO = \(aqL\(aq the lower trapezoid A(1:6,0:2) consists of the first
three columns of AP lower. The upper triangle A(0:2,0:2) consists of
the transpose of the last three columns of AP lower.
.br
This covers the case N even and TRANSR = \(aqN\(aq.
.br
       RFP A                   RFP A
.br
      03 04 05                33 43 53
.br
      13 14 15                00 44 54
.br
      23 24 25                10 11 55
.br
      33 34 35                20 21 22
.br
      00 44 45                30 31 32
.br
      01 11 55                40 41 42
.br
      02 12 22                50 51 52
.br
Now let TRANSR = \(aqT\(aq. RFP A in both UPLO cases is just the
transpose of RFP A above. One therefore gets:
.br
         RFP A                   RFP A
.br
   03 13 23 33 00 01 02    33 00 10 20 30 40 50
.br
   04 14 24 34 44 11 12    43 44 11 21 31 41 51
.br
   05 15 25 35 45 55 22    53 54 55 22 32 42 52
.br
We first consider Rectangular Full Packed (RFP) Format when N is
odd. We give an example where N = 5.
.br
   AP is Upper                 AP is Lower
.br
 00 01 02 03 04              00
.br
    11 12 13 14              10 11
.br
       22 23 24              20 21 22
.br
          33 34              30 31 32 33
.br
             44              40 41 42 43 44
.br
Let TRANSR = \(aqN\(aq. RFP holds AP as follows:
.br
For UPLO = \(aqU\(aq the upper trapezoid A(0:4,0:2) consists of the last
three columns of AP upper. The lower triangle A(3:4,0:1) consists of
the transpose of the first two columns of AP upper.
.br
For UPLO = \(aqL\(aq the lower trapezoid A(0:4,0:2) consists of the first
three columns of AP lower. The upper triangle A(0:1,1:2) consists of
the transpose of the last two columns of AP lower.
.br
This covers the case N odd and TRANSR = \(aqN\(aq.
.br
       RFP A                   RFP A
.br
      02 03 04                00 33 43
.br
      12 13 14                10 11 44
.br
      22 23 24                20 21 22
.br
      00 33 34                30 31 32
.br
      01 11 44                40 41 42
.br
Now let TRANSR = \(aqT\(aq. RFP A in both UPLO cases is just the
transpose of RFP A above. One therefore gets:
.br
         RFP A                   RFP A
.br
   02 12 22 00 01             00 10 20 30 40 50
.br
   03 13 23 33 11             33 11 21 31 41 51
.br
   04 14 24 34 44             43 44 22 32 42 52
.br
